# The number of different triangles and each side has a different length

I'm trying to solve the following problem:

What is the number of different triangles we can form from numbers $$4,5,6,7,8,9$$ (lengths of sides), where every side has a different length (for an example $$4,5,6$$ or $$4,5,7$$)..

My solution is the following:

The number of all possible permutations is $$\frac{6!}{(6-3)!}$$. We subtract the numbers which don't make a triangle (which is $$6*2$$) and then divide by $$2$$, because $$(4,5,6)$$ makes the same triangle is $$6,5,4$$. My answer is $$54$$.

However, the correct answer should be $$53$$. Can anyone tell me where I did a mistake?

Thanks

• $6 \choose 3$ is only $20$ and there is one choice $(4,5,9)$ that doesn't make a triangle, so I get $19$ Your expression (ignoring the factorial in the number to choose) gives $8$. Oct 20, 2019 at 16:42
• I think you'll have to show more details for someone to point out your mistake. What do you mean by $\binom{6}{(6-3)!}$? It look to me like ${6\choose6}=1$. Did you mean to say ${6\choose3}?$ Also, why do you say there are $12$ where the triangle inequality doesn't hold. How did you get that? Oct 20, 2019 at 16:43
• Sorry, I made a mistake when writing my answer here. Please wait a minute and I will fix it Oct 20, 2019 at 16:44
• If you write out all the triples $(4,5,6), (4,5,7), \ldots, (7,8,9)$, you'll see that there are only $20$ in the list, one of which is degenerate. How can there be $53$ or $54$ triangles? We could consider "right-handed" and "left-handed" triangles to be different, effectively doubling the number of triangles, but even then, we would have at most $40$, still falling short of the given answer. Have you written out the problem exactly as it was given to you? Oct 20, 2019 at 16:56
• You should have gotten six crossed off, as you describe. 2 for 4, 2 for 5 and 2 for 9 (not 6) makes six. Why do you double that? Then you should divide by $6$ because 456 comes six ways-456,465,546,564,654,645 not two Oct 20, 2019 at 17:04

There are $$\frac {6!}{3!}$$ ordered choices of three numbers, which is $$120$$. Six of those do not make a triangle, all the permutations of $$(4,5,9)$$, which leaves $$114$$. Each unordered triangle gives $$3!$$ permutations, so we divide by $$6$$ and get $$19$$. I don't know where a number in the $$50$$s comes from.
I think it is easier to just choose unordered combinations to start with, which is $${6 \choose 3}=20$$ and subtract the one that doesn't make a triangle. That also gets $$19$$.